Answer:
A third-degree polynomial can be written as:
f(x) = a*x^3 + b*x^2 + c*x + d
Where the leading coefficient is a, and all the coefficients are real.
If we know that the leading coefficient is 1, then the equation becomes:
f(x) = x^3 + b*x^2 + c*x + d
Now, we also know that:
(6 + i) and 5 are zeros.
This means that:
(6 + i)^3 + b*(6 + i)^2 + c*(6 + i) + d = 0
remember that:
i^2 = - 1
This is equal to:
(6 + i)*(36 + 2*6*i + i^2) + b*(36 + 2*6*i + i^2) + c*(6 + i) + d = 0
(6 + i)*(35 + 12i) + b*(35 + 12i) + c*(6 + i) + d =0
(210 + 35i + 72i - 12) + b*(35 + 12i) + c*(6 + i) + d = 0
198 + 107i + b*(35 + 12i) + c*(6 + i) + d = 0
sparating in real and imaginary part, we get:
(198 + b*35 + c*6 + d) + (107 + b*12 + c)*i = 0
Then each parentheses needs to be zero, this means that:
198 + b*35 + c*6 + d = 0
107 + b*12 + c = 0
Knowing that 5 is another zero, we have:
5^3 + b*5^2 + c*5 + d = 0
125 + b*25 + c*5 + d = 0
Then we have a system of 3 equations and 3 variables:
198 + b*35 + c*6 + d = 0
107 + b*12 + c = 0
125 + b*25 + c*5 + d = 0
To solve this, we first need to isolate one of the variables in one of the equations.
Let's isolate d in the last one, so we get:
d = -125 - b*25 - c*5
now we can replace this in the first equation to get:
198 + b*35 + c*6 + d = 0
198 + b*35 + c*6 + ( -125 - b*25 - c*5) = 0
70 + b*10 + c = 0
So now we have two equations:
70 + b*10 + c = 0
107 + b*12 + c = 0
Again, now we can isolate the one variable in one of the equations, this time let's isolate c in the first one.
c = -70 - b*10
now we can replace this in the other equation:
107 + b*12 + c = 0
107 + b*12 + (-70 - b*10) = 0
38 + b*2 = 0
now we can solve this for b
b*2 = -38
b = -38/2 = -19
Now, with the equation c = -70 - b*10 we can find the value of c.
c = -70 - b*10 = c = -70 - (-19)*10 = 120
And with the equation d = -125 - b*25 - c*5
we can find the value of d:
d = -125 - b*25 - c*5 = -125 - (-19)*25 - (120)*5 = -250
Then we have:
a = 1
b = -19
c = 120
d = -250
The eqation is:
f(x) = 1*x^3 - 19*x^2 + 120*x - 250
